It doesn't really make sense to say something is easy to see but you can't prove it. "It is easy to see this" is math jargon for "I have an easy proof of this".
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Carl MummertSep 12 '10 at 11:50

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@muad, anyone saying "it is easy to see" about the Jordan curve theorem only displays the fact that s/he has not thought about the thing enough and, more importantly, not even tried to actually prove it.
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Mariano Suárez-Alvarez♦Sep 12 '10 at 12:05

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It's not obvious, although the geometric model strongly suggests it when the curve is simple (e.g. a circle). In the present case, however, there is no geometric model that might suggest that no natural number has a non-integer rational square root, so the analogy is not very strong.
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Carl MummertSep 12 '10 at 12:21

One could say that the Jordan curve theorem seems obvious, but actually isn’t. One of the main aspects of mathematical training, I’d say, is learning to chase down your intuitions and either turn them into proofs (justifying that yes, something really was obvious!) or finding the weakness in the intuition (realising that actually something isn’t so obvious after all).
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Peter LeFanu LumsdaineNov 25 '10 at 22:06

11 Answers
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This is the second handout for a first course in number theory at the advanced undergraduate level. Three different proofs are discussed:

1) A generalization of the proof of irrationality of $\sqrt{2}$, using the decomposition of any positive integer into a perfect $k$th power times a $k$th power-free integer, followed by Euclid's Lemma. (For some reason, I don't give all the details of this proof. Maybe I should...)

2) A proof using the functions $\operatorname{ord}_p$, very much along the lines of the one Carl Mummert mentions in his answer.

3) A proof by establishing that the ring of integers is integrally closed. This is done directly from unique factorization, but afterwards I mention that it is a special case of the Rational Roots Theorem.

Let me also remark that every proof I have ever seen of this fact uses the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations) in some form. [Edit: I have now seen Robin Chapman's answer to the question, so this is no longer quite true.] However, if you want to prove any particular case of the result, you can use a brute force case-by-case analysis that avoids FTA.

The remark in your last paragraph is misleading since the results under discussion are much weaker than the property of being a UFD. Indeed, even the stronger property of being integrally closed is much weaker than UFD. One can deduce that Z is integrally closed from more general results that don't (immediately) imply that Z is a UFD.
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Bill DubuqueSep 12 '10 at 16:26

@BD: I certainly agree that for a general domain, being integrally closed is much weaker than being a UFD. But I'm having trouble thinking of an argument that one could give in an elementary number theory course which would show the integers are integrally closed but not that they are a UFD. What do you have in mind?
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Pete L. ClarkSep 12 '10 at 23:09

Why do you think that Robin's answer has any relevance to your remark? As I've argued at length elsewhere (via conductors) such descent-based proofs are just unwindings of ideal-theoretic proofs, so they really do invoke the Euclidean property or some essentially equivalent property of Z.
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Bill DubuqueSep 14 '10 at 2:06

From your handout, when $n$ is not a perfect $k$th power, $\sqrt[k]{n}\not\in\Bbb Q$. What happen with $\sqrt[k]{r}$ for rational and positive $r$?
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leoJun 11 '14 at 5:08

Unique prime factorisation implies that there exists a prime $p$ and positive integer $t$ such that $p^t$ divides $y$ while $p^t$ does not divide $x$. Therefore $p^{bt}$ divides $y^b$ while $p^{bt}$ does not divide $x^t$. Hence $y^b$ does not divide $x^b$.

[OOC: This answer has been through several revisions (some of the comments below might not relate to this version)]

The above Lemma is quite fundamental to factorization. I frequently refer to it by the suggestive moniker unique fractionization in order to highlight its equivalence to uniqueness of factorizations into irreducibles (one easily verifies that it is equivalent to Euclid's Lemma, which implies that irreducibles are prime). The structure implicit in the Lemma is a denominator or order ideal. Exploiting this structure, the proof easily generalizes to show that rational roots of monic integer coefficient polynomials must be integers, i.e. $\:\mathbb Z\:$ is integrally-closed (cf. the monic case of the Rational Root Test). In fact, this generalizes much further, employing Dedekind's key notion of a conductor ideal, to a one-line proof that PIDs are integrally closed. For much more on this see my post hereand especially the posts linked there, and their links $\ldots$ (it is a beautiful web of ideas - mostly all due to Dedekind - as Noether often rightly remarked).

As muad points out, you can also obtain this as an easy consequence of the Rational Root Theorem: if $a_nx^n+\cdots+a_0$ is a polynomial with integer coefficients, and $\frac{p}{q}$ is a rational root with $\gcd(p,q)=1$, then $p|a_0$ and $q|a_n$ (plug in, clear denominators, factor out).

So if you look at the polynomial $x^b-a$, with $b$ and $a$ positive integers, then a rational root must be of the form $\frac{p}{q}$, with $\gcd(p,q)=1$, and $q|1$. Thus, it must be an integer. So if it has a rational root, then the root is integral.

The general theorem is that a natural number $a$ has a rational square root if and only if the multiplicity of every prime factor of $a$ is even. For example $2^43^611^2$ has a rational square root but $5^411^3$ does not.

Moreover, if a natural number has a rational square root, that square root is always obtained by halving the multiplicity of each prime factor, and so the square root is also a natural number.

Right -- this argument can be stated very cleanly using the ord_p functions, and is one of the proofs I give in the notes I linked to in my answer. (This is probably my favorite proof, since it showcases the usefulness of the ord_p's.)
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Pete L. ClarkSep 12 '10 at 14:55

Below is a simple proof of irrationality of square-roots that I discovered as a teenager (inspired by a proof of Dedekind). It employs the Bezout identity for the gcd, i.e. the gcd $\rm\,(a,b)\,$ of integers $\rm\,a,b\,$ may be expressed as an integral linear combination of the given integers: $\rm\,\ (a,b)\, =\, a\, d - b\, c$

These are all essentially special cases of proofs that the ideal $\,a\Bbb Z + b\Bbb Z = 1\,$ using fast/slow descent based on the (Euclidean) Division algorithm (with remainder), i.e. specializations of the proof that ideals are principal in a Euclidean domain (see also various posts on denominator ideals).

That the numerator $\rm\,a = br\,$ is also a denom of $\rm\,r\,$ generalizes to any algebraic integer $\rm\,r,\,$ most efficiently by using Dedekind's conductor ideal. This generalizes the above proofs into a slick one-line proof that PIDs are integrally-closed, as I explained at length elsewhere. It beautifully abstracts the denominator descent that governs ad-hoc "elementary" irrationality proofs.

so that $(a-br)(c+dr)=ac-bdn+r$ and since $ac-bdn$ is an integer and $a-br=0$, we get that $r=-(ac-bdn)\in\mathbb{Z}$.
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robjohn♦Aug 18 '12 at 15:11

@Peter The $\rm\color{#C00}{leading}$ coef of the product, viewed as a polynomial in $\rm\:r,\:$ is $\rm\:ad\!-\!bc.\:$ We want this $ = \color{#C00}{\bf 1},$ since the goal of the proof is: $\,$ given that $\rm\:r\:$ is a root of a monic polynomial $\rm\:r^2-a = 0\:$ to deduce a lower-degree monic polynomial also having $\rm\:r\:$ as a root. The inductive step is clearer if you look at said higher-degree generalization. There, using the same degree-reduction step, by induction, we eventually reach a monic linear polynomial $\rm\:r - n = 0\:$ so $\rm\:r=n\in \Bbb Z.\:$
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Bill DubuqueAug 18 '12 at 15:34

To answer Pete's comment as to how to prove integral
closure of $\mathbb{Z}$ without using the UFD property.

Let $a/b$ be a rational ($a$, $b\in\mathbb{Z}$) which is
integral over $\mathbb{Z}$. Let $R=\mathbb{Z}[a/b]$. Then
$R$ is a finitely-generated $\mathbb{Z}$-module. It follows that
$b^n R\subseteq\mathbb{Z}$ for some $n$. We reduce to proving
the lemma that if $R$ is a ring with
$\mathbb{Z}\subseteq R\subseteq N^{-1}\mathbb{Z}$ for some
nonzero integer $N$ then $R=\mathbb{Z}$.

There are various ways of proving this. For instance if
$R\ne\mathbb{Z}$ there is an element of $R$ strictly between two
consecutive integers (this is the division algorithm) and so
an element $x$ of $R$ strictly between $0$ and $1$. If $y$ is the least
such number then considering $y^2$ gives a contradiction.
Alternatively, $M=xN$ is an integer and
$R\subseteq M^{-1}\mathbb{Z}$ so we can always replace $N$ by a smaller integer etc.

Thanks, Robin. I will note that you used the division algorithm, which is also what you use to prove that $\mathbb{Z}$ is a UFD. (Admittedly, if you hadn't said that you used it then I wouldn't have seen it.) I guess the question is: is this in any sense easier than just proving FTA? I'm not sure...
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Pete L. ClarkSep 13 '10 at 7:27

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You'll certainly need something like the division algorithm. The argument will work for Euclidean domains too. Note how easy it is to use the division algorithm without being aware of it :-)
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Robin ChapmanSep 13 '10 at 9:31

So... we can prove something pretty general here: the only rational roots of polynomials $x^n + \cdots + a_0$ with integer coefficients are integers.

Indeed, suppose $p/q$ is a rational root, in lowest form, then $p^n = -q(a_0q^{n-1} + a_1pq^{n-2} + \cdots + a_{n-1}p^{n-1})$. Now, if $q>1$, then any prime divisor of $q$ also divides $p^n$, and hence $p$. But this contradicts our assumption that $p/q$ is in lowest form, so we conclude that $q=1$, so the root is integral.

This is simply the monic case of the well-known rational root test which has already been mentioned above a few times. In more technical terms one simply says that Z is integrally closed (in its fraction field), i.e. fraction that is integral over Z (root of a monic polynomial) already lies in Z.
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Bill DubuqueSep 14 '10 at 1:53

I saw people mention the rational root test... but then they just applied it to polynomials like x^b-a ... the point is that the same works for monic polynomials in general. That's all :)
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Dylan WilsonSep 14 '10 at 1:55

They applied it to $ x^b - c $ simply because that's all that is required here. I don't recall anyone ever implying that it works only for such binomials.
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Bill DubuqueSep 14 '10 at 1:57

No worries, wasn't trying to be confrontational here. I'll remove my last sentence if it makes you feel better
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Dylan WilsonSep 14 '10 at 2:07

No problem, I was just trying to help ensure that you understood what was above.
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Bill DubuqueSep 14 '10 at 2:08

Proposition: For natural number $a$, $\sqrt{a}$ is irrational or an integer.

Proof: Suppose $\sqrt{a}$ is not an integer, but that it is rational. So $\sqrt{a} = \frac{p}{q}$ for some integer $p$ and natural number $q$. Thus $q \sqrt{a} = p$. That is, there exists at least one natural number $q$ such that $q \sqrt{a}$ is an integer. Let $m$ be the smallest natural number such that $m\sqrt{a}$ is an integer.

Since $\sqrt{a}$ is not an integer, there is an integer $b$ such that $b < \sqrt{a} < b+1$. Thus $0 < \sqrt{a} - b < 1$ and $0 < m(\sqrt{a} -b) < m$.

Consider the number $k = m(\sqrt{a} -b)$. First note that $k = m\sqrt{a} - mb$, which is an integer since both $m \sqrt{a}$ and $mb$ are integers.

Also, $k\sqrt{a} = (m \sqrt{a} - mb)\sqrt{a} = ma - bm\sqrt{a}$. Again this is an integer since $ma$, $b$ and $m\sqrt{a}$ are integers.

However, $0 < k < m$, so k is in fact a natural number less than $m$ such that $k\sqrt{a}$ is an integer.

This is a contradiction. So $\sqrt{a}$ cannot be rational if it is not an integer. $\blacksquare$

Proposition: For natural numbers $a$ and $t$, $a^{1/t}$ is either irrational or an integer.

Proof: Suppose $a^{1/t}$ is not an integer, but it is rational. Therefore all the integer powers of $a^{1/t}$ are also rational. Consider the list of rational numbers $a^{i/t}$ for $i = 1$, ..., $t-1$. These can be written with a single common denominator, and multiplying them all by this denominator will produce a list of integers. Let $m$ be the smallest natural number such that $ma^{i/t}$ is an integer for all of $i = 1$, ..., $t-1$. (We have already argued that such a number exists.)

Now $a^{1/t}$ is not an integer, so there exists an integer $b$ such that $b < a^{1/t} < b+1$. Then $0 < a^{1/t} - b < 1$ and so $0 < m(a^{1/t} -b) < m$. Consider the number $k = m(a^{1/t} - b)$.

Now $k = ma^{1/t} - mb$, which is an integer since $ma^{1/t}$ and $mb$ are integers.

Also, for any $i = 1$, ..., $t-1$, $k a^{i/t} = ma^{(i+1)/t} - bma^{i/t}$. This is an integer since $ma^{i/t}$ and $ma^{(i+1)/t}$ are both integers. (Note that for $i=t-1$, $a^{(t-1+1)/t} = a^{t/t} = a$, so $ma^{(i+1)/t}$ is still an integer for $i = t-1$.)

However, $0 < k < m$, so in fact $k$ is a natural number less than $m$ such that $ka^{i/t}$ is an integer for $i =1$, ..., $t-1$. $\blacksquare$

Note: this proof was inspired by "Irrationality without number theory", Richard Beigel, American Mathematical Monthly, 98 (1991) (doi - paywalled)